`(a+b+c)^2=3(ab+bc+ca)`

`<=>a^2+b^2+c^2+2ab+2bc+2ca=3(ab+bc+ca)`

`<=>a^2+b^2+c^2=ab+bc+ca`

`<=>2a^2+2b^2+2c^2=2ab+2bc+2ca`

`<=>(a-b)^2+(b-c)^2+(c-a)^2=0`

`VT>=0`

Dấu "=" xảy ra khi `a=b=c`

`a^3+b^3+c^3=3abc`

`<=>a^3+b^3+c^3-3abc=0`

`<=>(a+b)^3+c^3-3abc-3ab(a+b)=0`

`<=>(a+b)^3+c^3-3ab(a+b+c)=0`

`<=>(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0`

`**a+b+c=0`

`**a^2+b^2+c^2=ab+bc+ca`

`<=>a=b=c`

b) \(\left(a+b+c\right)^2=3\left(ab+bc+ca\right)\)

\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\) (chuyển vế qua)

\(\Leftrightarrow\frac{1}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)

Do VP >=0 với mọi a, b, c. Nên để đăng thức xảy ra thì a = b = c

c) a + b + c = 0 suy ra a = -(b+c)

\(a^3+b^3+c^3=b^3+c^3-\left(b+c\right)^3\)

\(=b^3+c^3-b^3-3bc\left(b+c\right)-c^3\)

\(=3bc.\left[-\left(b+c\right)\right]=3abc\) (đpcm)

a) \(a^2+b^2+c^2+3=2\left(a+b+c\right)\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)

Do VT >=0 với mọi a, b, c nên a = b = c 1

tí đăng tiếp

a/ Biến đổi tương đương:

\(\Leftrightarrow a^2c+ab^2+bc^2\ge b^2c+ac^2+a^2b\)

\(\Leftrightarrow a^2c-a^2b+ab^2-ac^2+bc^2-b^2c\ge0\)

\(\Leftrightarrow a^2\left(c-b\right)-\left(ab+ac\right)\left(c-b\right)+bc\left(c-b\right)\ge0\)

\(\Leftrightarrow\left(c-b\right)\left(a^2+bc-ab-ac\right)\ge0\)

\(\Leftrightarrow\left(c-b\right)\left(a\left(a-b\right)-c\left(a-b\right)\right)\ge0\)

\(\Leftrightarrow\left(c-b\right)\left(a-c\right)\left(a-b\right)\ge0\)

\(\Leftrightarrow\left(c-b\right)\left(c-a\right)\left(b-a\right)\ge0\) luôn đúng do \(a\le b\le c\)

Vậy BĐT ban đầu đúng

Câu 2: Đề sai, cho \(a=b=c=1\Rightarrow3\ge6\) (sai)

Đề đúng phải là \(\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

\(VT=\frac{a^2}{abc}+\frac{b^2}{abc}+\frac{c^2}{abc}=\frac{a^2+b^2+c^2}{abc}\ge\frac{ab+ac+bc}{abc}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

Câu 3: Không phải với mọi x; y với mọi \(x;y\) dương

Biến đổi tương đương do mẫu số vế phải dương nên ta được quyền nhân chéo:

\(\Leftrightarrow3x^3\ge\left(2x-y\right)\left(x^2+xy+y^2\right)\)

\(\Leftrightarrow3x^3\ge2x^3+x^2y+xy^2-y^3\)

\(\Leftrightarrow x^3+y^3-x^2y-xy^2\ge0\)

\(\Leftrightarrow x^2\left(x-y\right)-y^2\left(x-y\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)\left(x^2-y^2\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)^2\left(x+y\right)\ge0\) (luôn đúng)

3/ Áp dụng bất đẳng thức AM-GM, ta có :

\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\sqrt{\dfrac{\left(ab\right)^2}{\left(bc\right)^2}}=\dfrac{2a}{c}\)

\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2\sqrt{\dfrac{\left(bc\right)^2}{\left(ac\right)^2}}=\dfrac{2b}{a}\)

\(\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge2\sqrt{\dfrac{\left(ac\right)^2}{\left(ab\right)^2}}=\dfrac{2c}{b}\)

Cộng 3 vế của BĐT trên ta có :

\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)

\(\Leftrightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\left(\text{đpcm}\right)\)

Bài 1:

Áp dụng BĐT AM-GM ta có:

\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{1}{2\sqrt{a^2.bc}}+\frac{1}{2\sqrt{b^2.ac}}+\frac{1}{2\sqrt{c^2.ab}}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{2abc}\)

Tiếp tục áp dụng BĐT AM-GM:

\(\sqrt{bc}+\sqrt{ac}+\sqrt{ab}\leq \frac{b+c}{2}+\frac{c+a}{2}+\frac{a+b}{2}=a+b+c\)

Do đó:

\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2abc}\leq \frac{a+b+c}{2abc}\) (đpcm)

Dấu "=" xảy ra khi $a=b=c$

Bài 2:

Thay $1=a+b+c$ và áp dụng BĐT AM-GM ta có:

\(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=\frac{(a+1)(b+1)(c+1)}{abc}\)

\(=\frac{(a+a+b+c)(b+a+b+c)(c+a+b+c)}{abc}\)

\(\geq \frac{4\sqrt[4]{a.a.b.c}.4\sqrt[4]{b.a.b.c}.4\sqrt[4]{c.a.b.c}}{abc}=\frac{64abc}{abc}=64\)

Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$

a. \(a^3+a^2c-abc+b^2c+b^3\)

<=> \(\left(a^3+b^3\right)+c\left(a^2-ab+b^2\right)\)

<=> (\(\left(a+b\right)\left(a^2-ab+b^2\right)+c\left(a^2-ab+b^2\right)\)

<=> \(\left(a+b+c\right)\left(a^2-ab+b^2\right)\)

vì a+b+c =0 => đpcm

b. 2(a+1)(b+1)=(a+b)(a+b+2)

<=> \(2\left(ab+a+b+1\right)=\)\(a^2+ab+2a+ab+b^2+2b\)

<=> \(2ab+2a+2b+2=a^2ab+2a+ab+b^2+2b\)

<=> \(a^2+b^2=2\)=> đpcm

a. a^3+a^2c-abc+b^2c+b^3a3+a2c−abc+b2c+b3

<=> \left(a^3+b^3\right)+c\left(a^2-ab+b^2\right)(a3+b3)+c(a2−ab+b2)

<=> (\left(a+b\right)\left(a^2-ab+b^2\right)+c\left(a^2-ab+b^2\right)(a+b)(a2−ab+b2)+c(a2−ab+b2)

<=> \left(a+b+c\right)\left(a^2-ab+b^2\right)(a+b+c)(a2−ab+b2)

vì a+b+c =0 => đpcm

b. 2(a+1)(b+1)=(a+b)(a+b+2)

<=> 2\left(ab+a+b+1\right)=2(ab+a+b+1)=a^2+ab+2a+ab+b^2+2ba2+ab+2a+ab+b2+2b

<=> 2ab+2a+2b+2=a^2ab+2a+ab+b^2+2b2ab+2a+2b+2=a2ab+2a+ab+b2+2b

<=> a^2+b^2=2a2+b2=2=> đpcm

post ít một thôi

Ta có:

\(\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}\ge a+b+c\)

\(\Leftrightarrow\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}-a-b-c\ge0\)

\(\Leftrightarrow\frac{c^3-a^3}{a^2}+\frac{a^3-b^3}{b^2}+\frac{b^3-c^3}{c^2}\ge0\)

\(\Leftrightarrow\frac{c^5b^2-a^3b^2c^2+a^5c^2-b^3a^2c^2+b^5a^2-c^3a^2b^2}{a^2b^2c^2}\ge0\)

Dễ thấy: mẫu dương nên:

\(\frac{c^5b^2-a^3b^2c^2+a^5c^2-b^3a^2c^2+b^5a^2-c^3a^2b^2}{a^2b^2c^2}\ge0\)

\(\Leftrightarrow c^5b^2+a^5c^2+b^5a^2-a^2b^2c^2\left(a+b+c\right)\ge0\Leftrightarrow\)

\(\Leftrightarrow c^5b^2+a^5c^2+b^5a^2+c^5b^2+a^5c^2+b^5a^2-2a^2b^2c^2\left(a+b+c\right)\ge0\)

Chưa nghĩ ra tiếp :v

\(\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}\)

\(=\left(\frac{a^3}{b^2}+a\right)+\left(\frac{b^3}{c^2}+b\right)+\left(\frac{c^3}{a^2}+c\right)-a-b-c\)

Áp dụng BĐT AM-GM ta có:

\(\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}-a-b-c\ge2.\sqrt{\frac{a^3.a}{b^2}}+2.\sqrt{\frac{b^3.b}{c^2}}+2.\sqrt{\frac{c^3.c}{a^2}}-a-b-c\)\(=2\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)-a-b-c\)

Áp dụng BĐT Cauchy schwarz ta có: 

\(\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}-a-b-c\ge2.\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)-a-b-c\)\(\ge2\left[\frac{\left(a+b+c\right)^2}{a+b+c}\right]-a-b-c=2\left(a+b+c\right)-a-b-c=a+b+c\)

                                                                                                        (  đpcm )

Dấu " = " xảy ra \(\Leftrightarrow a=b=c\)